91影视

Write the expression for the time dilation.

$Y=\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$鈥︹�� (1)

Here, v is the speed of a twin, and c is the speed of light.

(a)

Substitute $v=\frac{4}{3}$in equation (1).

$$\begin{gathered} y=\frac{1}{\sqrt{1-{\displaystyle \frac{4^2}{5^2}}}} \\ =\frac{1}{\sqrt{1-{\displaystyle \frac{16}{25}}}} \\ =\sqrt{\frac{25}{9}} \\ =\frac{5}{3} \end{gathered}$$

As the time duration of one twin between birthday and birthday is 18 years, the time elapsed with respect to her twin brother will be,

$$\begin{gathered} T=纬(18years) \\ T=\left(\frac{5}{3}\right)脳\left(18years\right) \\ T=30years \end{gathered}$$

Hence, the age of a twin brother will be,

Age of a brother = 21 + 30 years

Age of a brother = 51 years

Therefore, the age of a twin brother is 51 years.

(b)

Write the equation to calculate the distance of star X.

d=vt 鈥︹�� (2)

Here, t is the time taken to reach star X, which is given as:

$t=\frac{T}{2}$

Substitute T=30 years in the above expression.

$$\begin{gathered} t=\frac{30}{2}years \\ =15years \end{gathered}$$

role="math" localid="1650589350709">Substitutev=45candt=15yearsinequation(2)d=45c脳15yearsd=4c脳3yearsd=12lightyears.

Therefore, the distance of star X is 12 years.

(c)

In frame S, write the x and t coordinates.

x = 12 light years = 12c years

t= 15 years

Hence, the coordinates will be,

(x,t) = (12c, 15) years

Therefore, the coordinates (x,t) of the jump is (12c,15) years.

(d)

The twin got on at the origin in $\stackrel{炉}{S}$, and at the time of walking along the road on , she鈥檚 still at the origin, so, the $\stackrel{炉}{x}$coordinate will be,

$\stackrel{炉}{x}=0$

Write the equation to calculate the $\stackrel{炉}{t}$coordinate.

$\stackrel{炉}{t}=\frac{t}{纬}$

Substitute t=15 years and $纬=\frac{5}{3}$in the above equation.

$$\begin{gathered} \stackrel{炉}{t}=\frac{15}{\left({\displaystyle \frac{5}{3}}\right)} \\ \stackrel{炉}{t}=15脳\frac{3}{5} \\ =9years \end{gathered}$$

Hence, the coordinates will be,

$\left(\stackrel{炉}{x},\stackrel{炉}{t}\right)=(0,9)years$

Therefore, the coordinates $\left(\stackrel{炉}{x},\stackrel{炉}{t}\right)$of the jump in $\stackrel{炉}{S}$ is (0,9) years.

(e)

Write the equation for the $\tilde{x}$coordinate in the frame$\tilde{S}$.

$$\begin{gathered} \tilde{x}=纬(x+vt) \\ Substitute12c=x,v=\frac{4}{5}c,纬=\frac{5}{3}andt15intheaboveequation. \\ \tilde{x}=\frac{5}{3}\left(12c+\left(\frac{4}{5}c\right)\left(15\right)\right) \\ =\frac{5}{3}(12c+12c) \\ =\frac{5}{3}\left(24c\right) \\ =40lightyears \end{gathered}$$

Write the equation for the $\tilde{t}$coordinate in frame $\tilde{S}$.

$\tilde{t}=纬\left(t+\left(\frac{v}{c^2}\right)x\right)$

$$\begin{gathered} Substitute12c=x,v=\frac{4}{5}c,纬\frac{5}{3}andt=15intheaboveequation \\ \tilde{t}=\frac{5}{3}\left(15+\left(\begin{array}{c}\frac{{\displaystyle \frac{4}{5}}c}{c^2}\\ \end{array}\right)\left(12c\right)\right) \\ =\frac{5}{3}(15+\left(\frac{4}{5}c脳\frac{1}{c^2}\right)12c) \\ =\frac{5}{3}(15+\frac{48}{5}) \\ onfurthersolving, \\ \tilde{t}=\frac{5}{3}\left(\frac{75+48}{5}\right)=\frac{5}{3}\left(\frac{123}{5}\right)=41years \end{gathered}$$

Hence, the coordinates will be, (40c, 41) years

Therefore, the coordinates$(\tilde{x},\widetilde{t)}$of the jump in $\tilde{S}$is (40c,41) years

(f)

Using the coordinate time from 9 years to 41 years, the return trip takes 32 years so, the timing in her watch when she gets home will be,

Time (when she gets home) = 41 years + 9 years

Time (When she gets home) = 50 years

Therefore, the reading of a watch when she gets home is 50 years .

(g)

(i)Just before she makes the jump:

Calculate the age of a brother just before she makes the jump.

$$\begin{gathered} t=\frac{\stackrel{炉}{t}}{纬} \\ t=\frac{9years}{\left({\displaystyle \frac{5}{3}}\right)} \\ t=5.4years \end{gathered}$$

As he started walking on their 21^stbirthday, i.e., at the age of 21 the age of a brother will be,

Age of the brother = (21 years)+(5.4 years)

Age of brother = 26.4 years

(ii)Just after she makes the jump:

Calculate the age of a brother just after she makes the jump.

$$\begin{gathered} t=\frac{\stackrel{炉}{t}}{纬} \\ t=\frac{41years}{\left({\displaystyle \frac{5}{3}}\right)} \\ t=(41years)脳\left(\frac{3}{5}\right) \\ t=24.6years \end{gathered}$$

As he started walking on their 21^stbirthday, i.e., at the age of 21 , the age of a brother will be,

Age of a brother will be = (21 years) + (24.6 years)

Age of a brother = 45.6 years

Therefore, (i) the age of a brother just before she makes the jump 26.4 years is and (ii) the age of a brother just after she makes the jump is 45.6 years.

(h)

It will take another of earth time for the return. Hence, the brother鈥檚 age will be,

Age of a brother = (45.6 years) + (5.4 years)

Age of a brother= 51 years

On comparing the above result to part (a), the age of a brother is exactly the same.

Therefore, the time required to return back is 5.4 years and the expected age of a brother at their reunion is 51 years.